Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{-9p + 90}{p^2 - 20p + 100} \times \dfrac{p^2 - 10p}{8p - 40} $
Solution: First factor the quadratic. $n = \dfrac{-9p + 90}{(p - 10)(p - 10)} \times \dfrac{p^2 - 10p}{8p - 40} $ Then factor out any other terms. $n = \dfrac{-9(p - 10)}{(p - 10)(p - 10)} \times \dfrac{p(p - 10)}{8(p - 5)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -9(p - 10) \times p(p - 10) } { (p - 10)(p - 10) \times 8(p - 5) } $ $n = \dfrac{ -9p(p - 10)(p - 10)}{ 8(p - 10)(p - 10)(p - 5)} $ Notice that $(p - 10)$ appears twice in both the numerator and denominator so we can cancel them. $n = \dfrac{ -9p(p - 10)\cancel{(p - 10)}}{ 8\cancel{(p - 10)}(p - 10)(p - 5)} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $n = \dfrac{ -9p\cancel{(p - 10)}\cancel{(p - 10)}}{ 8\cancel{(p - 10)}\cancel{(p - 10)}(p - 5)} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $n = \dfrac{-9p}{8(p - 5)} ; \space p \neq 10 $